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Q.

A body is released from the top of a smooth inclined plane of inclination $θ$ . It reaches the bottom with velocity $v$ . If the angle of inclination is doubled for the same length of the plane, what will be the velocity of the body on reaching the ground:

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a

$2 v$

b

${(2 \sin θ)}^{1 / 2} v$

c

${(2 \cos θ)}^{1 / 2} v$

d

$v$

answer is C.

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Detailed Solution

v= $\sqrt{2 gs (sinθ )}$

v′= $\sqrt{2 g s \sin 2 θ }$ = $\sqrt{2 g s (2 \sin θ \cos θ) }$

= $\sqrt{2 g s \sin θ }$ $\sqrt{2 \cos θ}$ = ${(2 \cos θ)}^{1 / 2} v$

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A body is released from the top of a smooth inclined plane of inclination θ. It reaches the bottom with velocity v . If the angle of inclination is doubled for the same length of the plane, what will be the velocity of the body on reaching the ground: